Find the root of a real-valued function with Newton's Method.

Example:
./parse.py "x^3-e^x"

Output:
['x', '^', '3', '-', 'e', '^', 'x']
('current token', 'x')
('current token', '^')
('current opr', '^')
('current token', '3')
('current token', '-')
('current opr', '-')
('22222222222', '-')
('current token', 'e')
('current token', '^')
('current opr', '^')
('current token', 'x')
(3, 2)
('before differention:', '(x^(3))-(2.71828182846^(x))')
fuck
('Differention:', '1*3*(x^(2))-ln(2.71828182846)*(2.71828182846^(x))*1')
('Simplify:', '3*(x^(2))-(2.71828182846^(x))')
Iterating times:1 value:89.000000
...
Iterating times:88 value:4.575997
Iterating times:89 value:4.537968
The root of the function is:4.537968